linux-kernel - Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE)

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Message-ID: <CABexPfF-31f1vcJpqdsDHKUHfes1230vwGMtm6ETZXVQQf4bcA@mail.gmail.com>
Date:	Wed, 23 Nov 2011 04:40:36 +0800
From:	zhihua che <zhihua.che@...il.com>
To:	linux-kernel@...r.kernel.org
Subject: Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE) - 1 in
 function slab_order()?

Hi, everyone,
        I'm reading the kernel codes about slub allocator and I come
across a confusion. Precisely, I'm reading the initialization of the
slub allocator, kmem_cache_init(), and I find it needs call
calculate_sizes() to determine the order of a kmem_cache, given the
size of the object. In turn, it calls the get_order() to get a
possible order. The problem is, in the start of this function, why it
looks like this:

        if (order_objects(min_order, size, reserved) > MAX_OBJS_PER_PAGE)
                return get_order(size * MAX_OBJS_PER_PAGE) - 1;

        I don't know why it subtracts one from the order returned by
get_order().
        because as far as I know, get_order() returns the order the
slab requires to reserve size * MAX_OBJS_PER_PAGE memory. If it
subtracts 1 from the order returned by get_order(), the slab can't
store MAX_OBJS_PER_PAGE objects at all, instead it can only store half
of the MAX_OBJS_PER_PAGE objects.
        Could you correct me if I think in a wrong way.
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